How do you simplify $sqrt(68)$ ?

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Answer 1 · 2016-05-23T21:49:13

$sqrt(68) = 2sqrt(17)$

If $a, b >= 0$ then:

$sqrt(ab) = sqrt(a)sqrt(b)$

The prime factorisation of $68$ is:

$68=2xx2xx17$

So we have:

$sqrt(68) = sqrt(2^2*17) = sqrt(2^2)*sqrt(17) = 2sqrt(17)$

$sqrt(17)$ cannot be simplified any further, but since $17=4^2+1$ is of the form $n^2+1$ it has a very simple continued fraction expansion:

$sqrt(17) = [4;bar(8)] = 4+1/(8+1/(8+1/(8+1/(8+1/(8+1/(8+...))))))$

You can terminate this expansion at any point to give a rational approximation. For example:

$sqrt(17) ~~ [4;8,8] = 4+1/(8+1/8) = 268/65 ~~ 4.1231$

How do you simplify sqrt(68)sqrt(68)?